### Work Power Energy Question 3

#### Question 3 - 25 January - Shift 1

An object of mass ’m’ initially at rest on a smooth horizontal plane starts moving under the action of force $F=2 N$. In the process of its linear motion, the angle $\theta$ (as shown in figure) between the direction of force and horizontal varies as $\theta=kx$, where $k$ is a constant and $x$ is the distance covered by the object from its initial position. The expression of kinetic energy of the object will be $E=\frac{n}{k} \sin \theta$. The value of $n$ is

Smooth horizontal surface

## Show Answer

#### Answer: (2)

#### Solution:

#### Formula: Kinetic Energy

$F \cos \theta=ma$

$2 \cos (kx)=\frac{mvdv}{dx}$

$\int_0^{v} v d v=2 \int_0^{x} \cos (k x) d x$

$\frac{mv^{2}}{2}=\frac{2}{k} \sin kx$

K.E. $=\frac{2}{k} \sin \theta$

$\mathbf{n}=\mathbf{2}$